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High School Physics : Electric Circuits

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Example Questions

Example Question #1 : Electricity And Magnetism

What is the voltage of a circuit with 10A $\displaystyle 10A$ of current and $20\Omega$ $\displaystyle 20\Omega$ of resistance?

Possible Answers:

30V $\displaystyle 30V$

10V $\displaystyle 10V$

0.5V $\displaystyle 0.5V$

200V $\displaystyle 200V$

$\displaystyle 2V$

Correct answer:

200V $\displaystyle 200V$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$. In this equation $\displaystyle V$ is the voltage, $\displaystyle I$ is the current, and $\displaystyle R$ is the resistance.

Plug in the given values and solve for the voltage.

$V=10A* 20\Omega$ $\displaystyle V=10A* 20\Omega$

V=200V $\displaystyle V=200V$

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Example Question #1 : Circuit Calculations And Concepts

What is the current of a circuit with a voltage of 12V $\displaystyle 12V$ and a total resistance of $5\Omega$ $\displaystyle 5\Omega$?

Possible Answers:

2.4A $\displaystyle 2.4A$

60A $\displaystyle 60A$

4.8A $\displaystyle 4.8A$

0.42A $\displaystyle 0.42A$

$\displaystyle 7A$

Correct answer:

2.4A $\displaystyle 2.4A$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ . In this equation $\displaystyle V$ is the voltage, $\displaystyle I$ is the current, and $\displaystyle R$ is the resistance.

We can re-arrange the equation to solve specifically for $\displaystyle I$.

$\frac{V}{R}=I$ $\displaystyle \frac{V}{R}=I$

Plug in the given values for voltage and resistance to solve for the current.

$\frac{V}{R}=I$ $\displaystyle \frac{V}{R}=I$

$\frac{12V}{5\Omega}=I$ $\displaystyle \frac{12V}{5\Omega}=I$

2.4A=I $\displaystyle 2.4A=I$

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Example Question #3 : Electricity And Magnetism

What is the resistance in a circuit with a voltage of 50V $\displaystyle 50V$ and a current of $\displaystyle 8A$?

Possible Answers:

$400\Omega$ $\displaystyle 400\Omega$

$58\Omega$ $\displaystyle 58\Omega$

$6.25\Omega$ $\displaystyle 6.25\Omega$

$0.16\Omega$ $\displaystyle 0.16\Omega$

$42\Omega$ $\displaystyle 42\Omega$

Correct answer:

$6.25\Omega$ $\displaystyle 6.25\Omega$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$ . In this equation $\displaystyle V$ is the voltage, $\displaystyle I$ is the current, and $\displaystyle R$ is the resistance.

We can re-arrange the equation to solve specifically for $\displaystyle R$.

$\frac{V}{I}=R$ $\displaystyle \frac{V}{I}=R$

Plug in the given values for voltage and current and solve for resistance.

$\frac{V}{I}=R$ $\displaystyle \frac{V}{I}=R$

$\frac{50V}{8A}=R$ $\displaystyle \frac{50V}{8A}=R$

$6.25\Omega=R$ $\displaystyle 6.25\Omega=R$

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Example Question #1 : Circuit Calculations And Concepts

A $\displaystyle 9V$ battery is connected to a circuit. The measured current is $\displaystyle 2A$. What is the equivalent resistance?

Possible Answers:

$4.5\Omega$ $\displaystyle 4.5\Omega$

$0.22\Omega$ $\displaystyle 0.22\Omega$

$10\Omega$ $\displaystyle 10\Omega$

$18\Omega$ $\displaystyle 18\Omega$

$9\Omega$ $\displaystyle 9\Omega$

Correct answer:

$4.5\Omega$ $\displaystyle 4.5\Omega$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$.

We are given the voltage and current, allowing us to solve for the resistance.

$\displaystyle 9V=2A* R$

$\frac{9V}{2A}=R$ $\displaystyle \frac{9V}{2A}=R$

$4.5\Omega=R$ $\displaystyle 4.5\Omega=R$

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Example Question #1 : Electricity And Magnetism

A closed electric circuit is set up so that there is a current of 9.9A $\displaystyle 9.9A$ and a voltage of 30.22V $\displaystyle 30.22V$. What is the resistance in the circuit?

Possible Answers:

$0.33\Omega$ $\displaystyle 0.33\Omega$

$3.05\Omega$ $\displaystyle 3.05\Omega$

$20.32\Omega$ $\displaystyle 20.32\Omega$

$299.18\Omega$ $\displaystyle 299.18\Omega$

$6.1\Omega$ $\displaystyle 6.1\Omega$

Correct answer:

$3.05\Omega$ $\displaystyle 3.05\Omega$

Explanation:

For this problem, use Ohm's law: V=IR $\displaystyle V=IR$.

We are given the current and the voltage. Using these terms, we can solve for the resistance.

V=IR $\displaystyle V=IR$

$\displaystyle 30.22V=(9.9A) R$

$\frac{30.22V}{9.9A}=R$ $\displaystyle \frac{30.22V}{9.9A}=R$

$3.05\Omega=R$ $\displaystyle 3.05\Omega=R$

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Example Question #4 : Electricity And Magnetism

The current in a circuit is 20A $\displaystyle 20A$. If the voltage is 35V $\displaystyle 35V$, what is the total resistance?

Possible Answers:

$700\Omega$ $\displaystyle 700\Omega$

$3.5\Omega$ $\displaystyle 3.5\Omega$

$0.57\Omega$ $\displaystyle 0.57\Omega$

$1.75\Omega$ $\displaystyle 1.75\Omega$

$70\Omega$ $\displaystyle 70\Omega$

Correct answer:

$1.75\Omega$ $\displaystyle 1.75\Omega$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the current and the voltage, allowing us to solve for the resistance.

$\displaystyle 35V=(20A)*R$

$R=\frac{35V}{20A}$ $\displaystyle R=\frac{35V}{20A}$

$R=1.75\Omega$ $\displaystyle R=1.75\Omega$

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Example Question #5 : Electricity And Magnetism

An electrical circuit has a total of $3.22\Omega$ $\displaystyle 3.22\Omega$ resistance and $\displaystyle 8V$. What is the current?

Possible Answers:

25.76A $\displaystyle 25.76A$

2.48A $\displaystyle 2.48A$

11.22A $\displaystyle 11.22A$

0.089A $\displaystyle 0.089A$

0.403A $\displaystyle 0.403A$

Correct answer:

2.48A $\displaystyle 2.48A$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the resistance and the voltage, allowing us to solve for the current.

$8V=I*(3.22\Omega)$ $\displaystyle 8V=I*(3.22\Omega)$

$I=\frac{8V}{3.22\Omega}$ $\displaystyle I=\frac{8V}{3.22\Omega}$

I=2.48A $\displaystyle I=2.48A$

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Example Question #6 : Electricity And Magnetism

An electrical circuit has a current of $\displaystyle 8A$ and $32\Omega$ $\displaystyle 32\Omega$ of resistance. What is the voltage?

Possible Answers:

0.25V $\displaystyle 0.25V$

$\displaystyle 4V$

256V $\displaystyle 256V$

128V $\displaystyle 128V$

40V $\displaystyle 40V$

Correct answer:

256V $\displaystyle 256V$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(8A)(32\Omega)$ $\displaystyle V=(8A)(32\Omega)$

V=256V $\displaystyle V=256V$

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Example Question #2 : Electricity And Magnetism

An electrical circuit has a current of 13A $\displaystyle 13A$ and $26\Omega$ $\displaystyle 26\Omega$ of resistance. What is the voltage?

Possible Answers:

338V $\displaystyle 338V$

$\displaystyle 2V$

13V $\displaystyle 13V$

0.5V $\displaystyle 0.5V$

39V $\displaystyle 39V$

Correct answer:

338V $\displaystyle 338V$

Explanation:

For this problem use Ohm's law:

V=IR $\displaystyle V=IR$

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(13A)(26\Omega)$ $\displaystyle V=(13A)(26\Omega)$

V=338V $\displaystyle V=338V$

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Example Question #2 : Electric Circuits

A circuit has a current, $\displaystyle I$, a voltage, $\displaystyle V$, and a resistance, $\displaystyle R$. If the voltage remains constant, but the current is doubled ($\displaystyle 2I$), what must the new resistance be?

Possible Answers:

$\frac{1}{2}R$ $\displaystyle \frac{1}{2}R$

$\displaystyle 2R$

There is insufficient information to solve

$\displaystyle 4R$

$\frac{1}{4}R$ $\displaystyle \frac{1}{4}R$

Correct answer:

$\frac{1}{2}R$ $\displaystyle \frac{1}{2}R$

Explanation:

To solve this problem, use Ohm's law:

V=IR $\displaystyle V=IR$

Since we are doubling the current, but the voltage is remaining the same, we can set our old and new equations equal to each other.

$\displaystyle I_1R_1=V=I_2R_2$

We know that the second current is equal to twice the first current.

$\displaystyle I_2=2I_1$

Use this equation to substitute current into the first equation.

$\displaystyle I_1R_1=(2I_1)R_2$

The initial current now cancels out from both sides.

$\displaystyle R_1=2R_2$

Divide both sides by two to isolate the final resistance variable.

$\frac{1}{2}R_1=R_2$ $\displaystyle \frac{1}{2}R_1=R_2$

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High School Physics : Electric Circuits

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Electricity And Magnetism

Example Question #1 : Circuit Calculations And Concepts

Example Question #3 : Electricity And Magnetism

Example Question #1 : Circuit Calculations And Concepts

Example Question #1 : Electricity And Magnetism

Example Question #4 : Electricity And Magnetism

Example Question #5 : Electricity And Magnetism

Example Question #6 : Electricity And Magnetism

Example Question #2 : Electricity And Magnetism

Example Question #2 : Electric Circuits

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